CODEWITHSUNDEEP
shell
Uvais Ibrahim
Uvais Ibrahim

Reputation: 741

searching specefic word in shell script

I have a problem.Please give me a solution.

I have to run a command as I given below, which will list all the files that contain the string given "abcde1234".

find /path/to/dir/ * | xargs grep abcde1234

But here it will display the files which contain the string "abcde1234567" also.But I nee only files which contain the word "abcde1234". What modification shall I need in the command ??

Upvotes: 0

Views: 390

Answers (3)

MangeshBiradar
MangeshBiradar

Reputation: 3938

$ grep abcde1234 *

This will grep the string abcde1234 in current directory, with the file name in which the string is. Ex:

abc.log: abcde1234 found

Upvotes: 1

Uvais Ibrahim
Uvais Ibrahim

Reputation: 741

Hi I got the answer for this.By attaching $ with word to be searched, it will display the files that contain only that word.

The command will be like this.

find /path/to/dir/ * | xargs grep abcde1234$

Thanks.

Upvotes: 0

aragaer
aragaer

Reputation: 17848

When I need something like that, I use the \< and \> which mean word boundary. Like this:

grep '\<abcde1234\>'

The symbols \< and \> respectively match the empty string at the beginning and end of a word.

But that's me. The correct way might be to use the -w switch instead (which I tend to forget about):

-w, --word-regexp
          Select  only  those  lines  containing matches that form whole words.  The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character.  Similarly, it
          must be either at the end of the line or followed by a non-word constituent character.  Word-constituent characters are letters, digits, and the underscore.

One more thing: instead of find + xargs you can use just find with -exec. Or actually just grep with -r:

grep -w -r abcde1234 /path/to/dir/

Upvotes: 4

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